Explaining this estimation of $\pi$ with an integral

Question

[This is potentially a duplicate] I was testing out integrals and I think I've found a way to evaluate a close approximation to $\pi$.

$$\int_{0}^{1} \dfrac{x^{4n}(1-x)^{4n}}{4^{n-1}(x^2 + 1)}\,dx$$

Where $n \in \mathbb{Z}^{+}$. I was testing this integral out on Wolfram and I noticed that the integral always contained a $\pm \pi$ as well as a fraction that was close to $0$. It seems like this is a way to find a fractional approximation to $\pi$.

Question: Why is this the case? I've seen the graph as $n$ becomes larger and the clearly approaches $0$, but why does this property occur?

My attempts at explaining it: By dividing the numerator by the denominator using polynomial long division for the first few values of $n$, you will end up with a polynomial with $\dfrac{4}{x^2+1}$ added to the end. I can see how $\pi$ is added to the end but I'm unsure how to prove this generally, this is only true for the first few values of $n$. I also tried to find a recurrence relation and find an expression from there, although I have been unsuccessful in my working. Any help or guidance in explaining this would be greatly appreciated!